Question

A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L of a 33.0 mM solution?

Answer 1

Answer:

5.41 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

$Moles=0.033 \times {0.45}\ moles$

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.01485\ g= \frac{Mass}{342.39\ g/mol}$

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

Answer -  5.41 g